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Pattern – 23 : Definite integration

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⋆ Pattern – 23

If f(t) is an odd function then

$g\left(x\right)\ =\ \int_{0}^{x}f\ \left(t\right)dt$ is an even function.

Proof :

$\therefore\ \ g\left(-x\right)=\ \int_{0}^{-x}f\left(t\right)dt$

Let, t = -y

∴ y = -t

∴ dt = -dy

t = 0 ⇒ y = 0

t = -x ⇒ y = x

$\therefore\ \ g\left(-x\right)=\ -\int_{0}^{x}f\left(-y\right)dy$

$=\ \int_{0}^{x}f\left(t\right)dt$

= g(x)

∴ g is an even function